7.4: Double-Angle, Half-Angle, and Reduction Formulas
- Page ID
- 114043
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section, you will:
- Use double-angle formulas to find exact values.
- Use double-angle formulas to verify identities.
- Use reduction formulas to simplify an expression.
- Use half-angle formulas to find exact values.
Figure 1 Bicycle ramps for advanced riders have a steeper incline than those designed for novices.
Bicycle ramps made for competition (see Figure 1) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be θθ such that tanθ=53.tanθ=53. The angle is divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one.
Using Double-Angle Formulas to Find Exact Values
In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where α=β.α=β. Deriving the double-angle formula for sine begins with the sum formula,
sin(α+β)=sinαcosβ+cosαsinβsin(α+β)=sinαcosβ+cosαsinβ
If we let α=β=θ,α=β=θ, then we have
sin(θ+θ)=sinθcosθ+cosθsinθ sin(2θ)=2sinθcosθsin(θ+θ)=sinθcosθ+cosθsinθ sin(2θ)=2sinθcosθ
Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, cos(α+β)=cosαcosβ−sinαsinβ,cos(α+β)=cosαcosβ−sinαsinβ, and letting α=β=θ,α=β=θ, we have
cos(θ+θ)=cosθcosθ−sinθsinθ cos(2θ)=cos2θ−sin2θcos(θ+θ)=cosθcosθ−sinθsinθ cos(2θ)=cos2θ−sin2θ
Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:
cos(2θ)=cos2θ−sin2θ =(1−sin2θ)−sin2θ =1−2sin2θcos(2θ)=cos2θ−sin2θ =(1−sin2θ)−sin2θ =1−2sin2θ
The second interpretation is:
cos(2θ)=cos2θ−sin2θ =cos2θ−(1−cos2θ) =2cos2θ−1cos(2θ)=cos2θ−sin2θ =cos2θ−(1−cos2θ) =2cos2θ−1
Similarly, to derive the double-angle formula for tangent, replacing α=β=θα=β=θ in the sum formula gives
tan(α+β)=tanα+tanβ1−tanαtanβtan(θ+θ)=tanθ+tanθ1−tanθtanθtan(2θ)=2tanθ1−tan2θtan(α+β)=tanα+tanβ1−tanαtanβtan(θ+θ)=tanθ+tanθ1−tanθtanθtan(2θ)=2tanθ1−tan2θ
The double-angle formulas are summarized as follows:
sin(2θ)=2sinθcosθsin(2θ)=2sinθcosθ
cos(2θ)=cos2θ−sin2θ =1−2sin2θ =2cos2θ−1cos(2θ)=cos2θ−sin2θ =1−2sin2θ =2cos2θ−1
tan(2θ)=2tanθ1−tan2θtan(2θ)=2tanθ1−tan2θ
Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.
- Draw a triangle to reflect the given information.
- Determine the correct double-angle formula.
- Substitute values into the formula based on the triangle.
- Simplify.
EXAMPLE 1
Using a Double-Angle Formula to Find the Exact Value Involving Tangent
Given that tanθ=−34tanθ=−34 and θθ is in quadrant II, find the following:
- ⓐ sin(2θ)sin(2θ)
- ⓑ cos(2θ)cos(2θ)
- ⓒ tan(2θ)tan(2θ)
- Answer
-
Given sinα=58,sinα=58, with θθ in quadrant I, find cos(2α).cos(2α).
EXAMPLE 2
Using the Double-Angle Formula for Cosine without Exact Values
Use the double-angle formula for cosine to write cos(6x)cos(6x) in terms of cos(3x).cos(3x).
- Answer
-
Analysis
This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.
Using Double-Angle Formulas to Verify Identities
Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.
EXAMPLE 3
Using the Double-Angle Formulas to Establish an Identity
Establish the following identity using double-angle formulas:
1+sin(2θ)=(sinθ+cosθ)21+sin(2θ)=(sinθ+cosθ)2
- Answer
-
Analysis
This process is not complicated, as long as we recall the perfect square formula from algebra:
(a±b)2=a2±2ab+b2(a±b)2=a2±2ab+b2
where a=sinθa=sinθ and b=cosθ.b=cosθ. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.
Establish the identity: cos4θ−sin4θ=cos(2θ).cos4θ−sin4θ=cos(2θ).
EXAMPLE 4
Verifying a Double-Angle Identity for Tangent
Verify the identity:
tan(2θ)=2cotθ−tanθtan(2θ)=2cotθ−tanθ
- Answer
-
Analysis
Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show
2tanθ1−tan2θ=2cotθ−tanθ2tanθ1−tan2θ=2cotθ−tanθ
Let’s work on the right side.
2cotθ−tanθ=21tanθ−tanθ(tanθtanθ) =2tanθ1tanθ(tanθ)−tanθ(tanθ) =2tanθ1−tan2θ2cotθ−tanθ=21tanθ−tanθ(tanθtanθ) =2tanθ1tanθ(tanθ)−tanθ(tanθ) =2tanθ1−tan2θ
When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.
Verify the identity: cos(2θ)cosθ=cos3θ−cosθsin2θ.cos(2θ)cosθ=cos3θ−cosθsin2θ.
Use Reduction Formulas to Simplify an Expression
The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.
We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with cos(2θ)=1−2sin2θ.cos(2θ)=1−2sin2θ. Solve for sin2θ:sin2θ:
cos(2θ)=1−2sin2θ2sin2θ=1−cos(2θ) sin2θ=1−cos(2θ)2cos(2θ)=1−2sin2θ2sin2θ=1−cos(2θ) sin2θ=1−cos(2θ)2
Next, we use the formula cos(2θ)=2cos2θ−1.cos(2θ)=2cos2θ−1. Solve for cos2θ:cos2θ:
cos(2θ)=2cos2θ−11+cos(2θ)=2cos2θ1+cos(2θ)2=cos2θ cos(2θ)=2cos2θ−11+cos(2θ)=2cos2θ1+cos(2θ)2=cos2θ
The last reduction formula is derived by writing tangent in terms of sine and cosine:
tan2θ=sin2θcos2θ =1−cos(2θ)21+cos(2θ)2 =(1−cos(2θ)2)(21+cos(2θ)) =1−cos(2θ)1+cos(2θ)Substitute the reduction formulas.tan2θ=sin2θcos2θ =1−cos(2θ)21+cos(2θ)2Substitute the reduction formulas. =(1−cos(2θ)2)(21+cos(2θ)) =1−cos(2θ)1+cos(2θ)
The reduction formulas are summarized as follows:
sin2θ=1−cos(2θ)2sin2θ=1−cos(2θ)2
cos2θ=1+cos(2θ)2cos2θ=1+cos(2θ)2
tan2θ=1−cos(2θ)1+cos(2θ)tan2θ=1−cos(2θ)1+cos(2θ)
EXAMPLE 5
Writing an Equivalent Expression Not Containing Powers Greater Than 1
Write an equivalent expression for cos4xcos4x that does not involve any powers of sine or cosine greater than 1.
- Answer
-
Analysis
The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.
EXAMPLE 6
Using the Power-Reducing Formulas to Prove an Identity
Use the power-reducing formulas to prove
sin3(2x)=[12sin(2x)][1−cos(4x)]sin3(2x)=[ 12sin(2x) ][ 1−cos(4x) ]
- Answer
-
Analysis
Note that in this example, we substituted
1−cos(4x)21−cos(4x)2
for sin2(2x).sin2(2x). The formula states
sin2θ=1−cos(2θ)2sin2θ=1−cos(2θ)2
We let θ=2x,θ=2x, so 2θ=4x.2θ=4x.
Use the power-reducing formulas to prove that 10cos4x=154+5cos(2x)+54cos(4x).10cos4x=154+5cos(2x)+54cos(4x).
Using Half-Angle Formulas to Find Exact Values
The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace θθ with α2,α2, the half-angle formula for sine is found by simplifying the equation and solving for sin(α2).sin(α2). Note that the half-angle formulas are preceded by a ±± sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which α2α2 terminates.
The half-angle formula for sine is derived as follows:
sin2θ=1−cos(2θ)2sin2(α2)=1−cos(2⋅α2)2=1−cosα2sin(α2)=±1−cosα2−−−−−−√ sin2θ=1−cos(2θ)2sin2(α2)=1−cos(2⋅α2)2=1−cosα2sin(α2)=±1−cosα2
To derive the half-angle formula for cosine, we have
cos2θ=1+cos(2θ)2cos2(α2)=1+cos(2⋅α2)2 =1+cosα2 cos(α2)=±1+cosα2−−−−−−√ cos2θ=1+cos(2θ)2cos2(α2)=1+cos(2⋅α2)2 =1+cosα2 cos(α2)=±1+cosα2
For the tangent identity, we have
tan2θ=1−cos(2θ)1+cos(2θ)tan2(α2)=1−cos(2⋅α2)1+cos(2⋅α2) =1−cosα1+cosα tan(α2)=±1−cosα1+cosα−−−−−−√ tan2θ=1−cos(2θ)1+cos(2θ)tan2(α2)=1−cos(2⋅α2)1+cos(2⋅α2) =1−cosα1+cosα tan(α2)=±1−cosα1+cosα
The half-angle formulas are as follows:
sin(α2)=±1−cosα2−−−−−−−−√sin(α2)=±1−cosα2
cos(α2)=±1+cosα2−−−−−−−−√cos(α2)=±1+cosα2
tan(α2)=±1−cosα1+cosα−−−−−−√=sinα1+cosα=1−cosαsinαtan(α2)=±1−cosα1+cosα=sinα1+cosα=1−cosαsinα
EXAMPLE 7
Using a Half-Angle Formula to Find the Exact Value of a Sine Function
Find sin(15∘)sin(15∘) using a half-angle formula.
- Answer
-
Analysis
Notice that we used only the positive root because sin(15o)sin(15o) is positive.
Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.
- Draw a triangle to represent the given information.
- Determine the correct half-angle formula.
- Substitute values into the formula based on the triangle.
- Simplify.
EXAMPLE 8
Finding Exact Values Using Half-Angle Identities
Given that tanα=815tanα=815 and αα lies in quadrant III, find the exact value of the following:
- ⓐ sin(α2)sin(α2)
- ⓑ cos(α2)cos(α2)
- ⓒ tan(α2)tan(α2)
- Answer
-
Given that sinα=−45sinα=−45 and αα lies in quadrant IV, find the exact value of cos(α2).cos(α2).
EXAMPLE 9
Finding the Measurement of a Half Angle
Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of θθ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If tanθ=53tanθ=53 for higher-level competition, what is the measurement of the angle for novice competition?
- Answer
-
Access these online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas.
7.3 Section Exercises
Verbal
1.
Explain how to determine the reduction identities from the double-angle identity cos(2x)=cos2x−sin2x.cos(2x)=cos2x−sin2x.
2.
Explain how to determine the double-angle formula for tan(2x)tan(2x) using the double-angle formulas for cos(2x)cos(2x) and sin(2x).sin(2x).
3.
We can determine the half-angle formula for tan(x2)=1−cosx√1+cosx√tan(x2)=1−cosx1+cosx by dividing the formula for sin(x2)sin(x2) by cos(x2).cos(x2). Explain how to determine two formulas for tan(x2)tan(x2) that do not involve any square roots.
4.
For the half-angle formula given in the previous exercise for tan(x2),tan(x2), explain why dividing by 0 is not a concern. (Hint: examine the values of cosxcosx necessary for the denominator to be 0.)
Algebraic
For the following exercises, find the exact values of a) sin(2x),sin(2x), b) cos(2x),cos(2x), and c) tan(2x)tan(2x) without solving for x.x.
5.
If sinx=18,sinx=18, and xx is in quadrant I.
6.
If cosx=23,cosx=23, and xx is in quadrant I.
7.
If cosx=−12,cosx=−12, and xx is in quadrant III.
8.
If tanx=−8,tanx=−8, and xx is in quadrant IV.
For the following exercises, find the values of the six trigonometric functions if the conditions provided hold.
9.
cos(2θ)=35cos(2θ)=35 and 90∘≤θ≤180∘90∘≤θ≤180∘
10.
cos(2θ)=12√cos(2θ)=12 and 180∘≤θ≤270∘180∘≤θ≤270∘
For the following exercises, simplify to one trigonometric expression.
11.
2sin(π4)2cos(π4)2sin(π4)2cos(π4)
12.
4sin(π8)cos(π8)4sin(π8)cos(π8)
For the following exercises, find the exact value using half-angle formulas.
13.
sin(π8)sin(π8)
14.
cos(−11π12)cos(−11π12)
15.
sin(11π12)sin(11π12)
16.
cos(7π8)cos(7π8)
17.
tan(5π12)tan(5π12)
18.
tan(−3π12)tan(−3π12)
19.
tan(−3π8)tan(−3π8)
For the following exercises, find the exact values of a) sin(x2),sin(x2), b) cos(x2),cos(x2), and c) tan(x2)tan(x2) without solving for x,x, when 0∘≤x≤360∘0∘≤x≤360∘
20.
If tanx=−43,tanx=−43, and xx is in quadrant IV.
21.
If sinx=−1213,sinx=−1213, and xx is in quadrant III.
22.
If cscx=7,cscx=7, and xx is in quadrant II.
23.
If secx=−4,secx=−4, and xx is in quadrant II.
For the following exercises, use Figure 5 to find the requested half and double angles.
Figure 5
24.
Find sin(2θ),cos(2θ),sin(2θ),cos(2θ), and tan(2θ).tan(2θ).
25.
Find sin(2α),cos(2α),sin(2α),cos(2α), and tan(2α).tan(2α).
26.
Find sin(θ2),cos(θ2),sin(θ2),cos(θ2), and tan(θ2).tan(θ2).
27.
Find sin(α2),cos(α2),sin(α2),cos(α2), and tan(α2).tan(α2).
For the following exercises, simplify each expression. Do not evaluate.
28.
cos2(28∘)−sin2(28∘)cos2(28∘)−sin2(28∘)
29.
2cos2(37∘)−12cos2(37∘)−1
30.
1−2sin2(17∘)1−2sin2(17∘)
31.
cos2(9x)−sin2(9x)cos2(9x)−sin2(9x)
32.
4sin(8x)cos(8x)4sin(8x)cos(8x)
33.
6sin(5x)cos(5x)6sin(5x)cos(5x)
For the following exercises, prove the identity given.
34.
(sint−cost)2=1−sin(2t)(sint−cost)2=1−sin(2t)
35.
sin(2x)=−2sin(−x)cos(−x)sin(2x)=−2sin(−x)cos(−x)
36.
cotx−tanx=2cot(2x)cotx−tanx=2cot(2x)
37.
1+cos(2θ)sin(2θ)tan2θ=tanθ1+cos(2θ)sin(2θ)tan2θ=tanθ
For the following exercises, rewrite the expression with an exponent no higher than 1.
38.
cos2(5x)cos2(5x)
39.
cos2(6x)cos2(6x)
40.
sin4(8x)sin4(8x)
41.
sin4(3x)sin4(3x)
42.
cos2xsin4xcos2xsin4x
43.
cos4xsin2xcos4xsin2x
44.
tan2xsin2xtan2xsin2x
Technology
For the following exercises, reduce the equations to powers of one, and then check the answer graphically.
45.
tan4xtan4x
46.
sin2(2x)sin2(2x)
47.
sin2xcos2xsin2xcos2x
48.
tan2xsinxtan2xsinx
49.
tan4xcos2xtan4xcos2x
50.
cos2xsin(2x)cos2xsin(2x)
51.
cos2(2x)sinxcos2(2x)sinx
52.
tan2(x2)sinxtan2(x2)sinx
For the following exercises, algebraically find an equivalent function, only in terms of sinxsinx and/or cosx,cosx, and then check the answer by graphing both equations.
53.
sin(4x)sin(4x)
54.
cos(4x)cos(4x)
Extensions
For the following exercises, prove the identities.
55.
sin(2x)=2tanx1+tan2xsin(2x)=2tanx1+tan2x
56.
cos(2α)=1−tan2α1+tan2αcos(2α)=1−tan2α1+tan2α
57.
tan(2x)=2sinxcosx2cos2x−1tan(2x)=2sinxcosx2cos2x−1
58.
(sin2x−1)2=cos(2x)+sin4x(sin2x−1)2=cos(2x)+sin4x
59.
sin(3x)=3sinxcos2x−sin3xsin(3x)=3sinxcos2x−sin3x
60.
cos(3x)=cos3x−3sin2xcosxcos(3x)=cos3x−3sin2xcosx
61.
1+cos(2t)sin(2t)−cost=2cost2sint−11+cos(2t)sin(2t)−cost=2cost2sint−1
62.
sin(16x)=16sinxcosxcos(2x)cos(4x)cos(8x)sin(16x)=16sinxcosxcos(2x)cos(4x)cos(8x)
63.
cos(16x)=(cos2(4x)−sin2(4x)−sin(8x))(cos2(4x)−sin2(4x)+sin(8x))